# Problem 14.4b1.1

Closed string degeneracies | A First Course in String Theory

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(b) State the values of $\alpha' M^2$ and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

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Type IIA closed superstrings

p.322

In closed superstring theories spacetime bosons arise from the (NS, NS) sector and also from the (R, R) sector, since this sector is “doubly” fermionic. The spacetime fermions arise from the (NS, R) and (R, NS) sectors.

p.322 $\alpha' M_L^2 = \alpha' M_R^2$

 $\frac{1}{2} \alpha' M^2 =$ $\alpha' M_L^2 + \alpha' M_R^2$ $\alpha' M^2 =$ $2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right)$ $=$ $4 \alpha' M_L^2$

Equation (14.77):

$\text{type IIA}:~~~(NS+, NS+), ~(NS+, R+),~ (R-, NS+), ~ (R-, R+)$

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What is the difference of the meanings of R+ and R-?

$R+$ states are world-sheet bosonic states.

p.316

It thus follows that all eight $| R_a \rangle$ states are fermionic and all $| R_{\bar a} \rangle$ are bosonic.

Be careful:

Here, “fermionic”/”bosonic” refers to the world-sheet fermions/bosons, not the spacetime ones.

p.320

Identifying $| R_a \rangle$ as spacetime fermions and $| R_{\bar a} \rangle$ as spacetime bosons is not an alternative either, since spacetime bosons cannot carry a spinor index.

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How come the R+ cannot be the left-moving part?

p.320

A strategy then emerges. Since all states in the R sector have a spinor index, we will only attempt to get spacetime fermions from this sector. We also recognize that all fermions must arise from states with the same value of $(-1)^F$.

Me@2015.09.11 10:36 AM: In other words, it is a convention:

Following Gliozzi, Scherk, and Olive (GSO) we proceed to truncate the Ramond sector down to the set of states with $(-1)^F = -1$.

— Me@2018-05-01 05:59:53 PM

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